Polynomials which permute matrices over commutative antinegative semirings

Polynomials Which Permute Matrices Over Commutative Antinegative Semirings* LeRoy B. Beasley Utah State University Logan, Utah 84322-3900 and Norman J. Pullman

Queen’s University Kingston, Ontario, Canada K7L 3N6

Submitted by Richard A. Brualdi

ABSTRACT Suppose

S is any commutative,

reals, the nonnegative

integers,

show that when n > 1, the operator surjective

antinegative

semiring

(such

or a Boolean algebra) and a,, a,,

as the nonegative

. . . , a, are in §. We

X + ~~=a=oakXk on the n X n matrices

over S is

if and only if a, is a unit and all other ak = 0.

In a series of papers Gilmer, G. E. Schnibben, conditions

[l-5], J. V. Brawley and coauthors and T. Vaughan

studied necessary

L. Carliz, R. and sufficient

for a matrix operator

X + a,Z + a,X

+

. . . + a,Xm

induced by a scalar polynomial p(r) = Cy=,,ajxj to be bijective on the n X n matrices over various rings and fields. For example, in [5], Brawley and Schnibben proved that a polynomial with coefficients in an algebraically

*This work was supported, in part, by the Natural Sciences and Engineering

Research

Council of Canada, under grant 0GPOOO4041.

LINEAR

AZGEBZU

AND ITS APPLICATIONS

165:167-172

(1992)

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0 Elsevier Science Publishing Co., Inc., 1992 655 Avenue of the Americas, New York, NY 10010

0024-3795/92/$5.00

168

LEROY B. BEASLEY AND NORMAN J. PULLMAN

closed field [F of characteristic

0 induces

a bijective

matrix operator

on Fn,,,,

the ring of n X n matrices over [F, if and only if p(r) is linear. Recently, Chuang [6] has characterized the sets of matrices which are the range of a polynomial over a finite field. He showed that the subset, M, of F” “, where IF is GF(q), is the range of a polynomial in noncommuting indeterminates with constant term 0 if and only if 0 E M and Bs’B-’ c & for all invertible Z3E FPl,,,. In this paper we will consider the problem of characterizing those polynomials with coefficients in S which are surjective on S,,,, where S is an antinegative semiring. A semiring (see e.g. [7] or 181) has the same axioms as a ring with unity, except that it is only assumed to be an abelian monoid under addition; it may not be closed under subtraction. Indeed, when only the additive identity, 0, has a negative, the semiring is said to be antinegative. Such structures occur naturally in combinatorial settings. For example, the nonnegative reals, the nonnegative integers, the fuzzy scalars, and the Boolean algebras are all antinegative semirings. In the Boolean case, addition is union, multiplication is intersection, 0 is 0, and the multiplicative identity 1, is the complement of 0. These examples are all commutative. Using the usual definitions of addition and multiplication of matrices, &,(S), the set of n X n matrices over any antinegative semiring S, is also an antinegative semiring; it is not commutative when n > 1. The n X n zero matrix, O,, and the n X n identity matrix, I,, are defined over semirings as they are over rings; their subscripts may be omitted if they can be inferred from the context. They are the additive

and multiplicative identities of J&(S). The notions of unit and are defined for semirings in the same way they are defined for rings. We will prove that when n > 1 and S is a commutative, antinegative

divisor

semiring,

C~+ajxj

the matrix operator on kn(S) induced by the polynomial p(x) = is surjective if and only if a, is a unit and all other coefficients

are 0. When

n = 1, the characterization

of surjectivity

depends on the semiring.

For example, over the nonnegative reals, p(r) is surjective if and only if a o = 0.Over the nonnegative integers, p(x) is surjective if and only if a, = 1 and aj = 0 for all j # 1. Over any Boolean algebra, p(x) is surjective if and only if a, = 0 and Cj”_laj = 1. Before we prove our characterization theorem, we need some additional notation and a few observations. Hereafter, S is a commutative, antinegative semiring and .k = kn(S), where n > 1. We also use the notation A@ B to denote the direct sum of matrices A and B. For X,Y in .k, we say X is dominated by Y, written X < Y or Y 2 X, if for some 2 in .k, X + 2 = Y. The antinegativity of S is inherited by .L, so

MATRIX-PERMUTING

169

POLYNOMIALS

X+Y=OifandonlyifX=Y=0;thus,X~OifandonlyifX=O.Italso follows that aX < aY for all a in S when X < Y.

LEMMA1. Zf p(x) = Ckm,,,ukxk is a polynomial over s, then the matrix operator on .A? induced by p is surjective only if a, = 0.

~‘roof.

If the operator

induced

by p is surjective,

there is some Y in .4? such that a,Z + CrSiakYk tive, a,Z = 0, and hence a, = 0.

then, in particular,

= 0. Since

.A? is antinegaW

For X in k, let xi:’ denote the (i,j)th entry in Xk. It follows, directly from the definitions of matrix multiplication and antinegativity, that for all indices

i, j, k, 1, and r,

X!k’X(‘! < x!‘r+‘) tr r, :.I ’

(1)

The following lemma is readily established inductively, definition of matrix multiplication and the inequality (1).

LEMMA2. (a) b,,

Suppose

directly

from the

matrix over C5 Then for all k > 1,

B is any 2X2

divides b,,(k), < bgk), and = 0, then b,, divides bi;k-l)

(b) (bzlblz)k

(c) $b,,

and (b,,b,,)kb,,

LEMMA3. Zf p(X) = Cpz,akxk is a polynomial is a 2 X2 matrix B over s such that

then a, is a unit and ak = 0 for all k > 1.

Proof.

Let

WC

1 1

[1

0’1

< bgkk+‘).

over s, m 2 1, and there

170 Since

LEROY B. BEASLEY AND NORMAN J. PULLMAN p(B) = W, we have, for all i and j,

5 a,b$).

wij =

(2)

k=l

First

we’ll show that b,, and b,, are units. By Lemma 2(a), for each exists an ok in s such that b’,k,)= bleak. Since wra = 1,

k > 1, there

Equation (2) implies that Symmetrically, so is b,,. Let u denote the unit imply that for all k 2 1,

1 = b12C;IzIak. and hence Then

b,,b,,.

bczk+‘) > b,,uk 22

Lemma

and

that

b,,

2(b) and the inequality

b$fzk)> uk.

Since wz2 = 0, Equation (2) and antinegativity imply that a,b,, also imply, via the inequalities (3) that for all k > 1, a2k_lb22

= 0

and

is a unit.

azk = 0.

(1)

(3)

= 0. They

(4)

Since wr2 = 1, Equation (2) implies b,, = cF= lak b,,b’,k,‘. Therefore, b,, = 0 by (4). By the inequality (1) and Lemma 2(c), we have bgk+‘) > ukb,,. Since wz2 = 0, Equation (2) and antinegativity imply that a2k+lukb,, = 0 and hence, for all k > 1,

blla2k+l

According

to Equations

=

O.

(2) and (4),

1 = a,b,,

+

E

a2k+lbgk+1)

k=l

because

(5)

w 11 = 1. By Lemma

(6)

2(c), for some & E s and all k 2 1,

(7) Then (5) (6) and (7) imply that a, b,, = 1. Consequently a, and b,, are units. Therefore, oak + r = 0 for all k 2 1, by Equation (5). We have already seen in (4) that ask = 0 for all k > 1. That completes the proof of the lemma. n

MATRIX-PERMUTING

171

POLYNOMIALS

THEOREM. lf n > 1, S is any commutative, antinegative semiring, and p(x) = Cr=‘,,a,x k is any polynomial whose coefficients are in S, then the operator on the n x n matrices induced by p is surjective if and only if a, is a unit and ak = 0 fm all k # 1.

Proof. The sufficiency is immediate, so we turn to a proof necessity. Suppose the operator X + CT=aakXk is surjective. Then by Lemma 1. Let

of the

a, = 0

1 1

WC

[1

01

and Z= W@O,_,, the direct sum of W and the (n-2)X(n-2) matrix. By surjectivity, there is a matrix X such that for all i and j,

zero

m

zij

c

=

akx$‘.

(8)

k=l

We shall show that X = B@C for some 2x2 matrix B and (n -2)X(n matrix C, and hence that p(B) = W. The th eorem then follows directly Lemma 2.

-2) from

We have

(k)zZ

akx12

-1

12-

k=l

)

(9)

and hence, m xi1

=

C akxilx$).

(10)

k=l

Suppose i > 2. We shall prove that xi1 = 0, by showing that each term tk = akxilx(li) in (10) is 0. If k 2 2, then tk = ~akxilxlql(n~::xq 4 >xqt_lz, I 1+1 where the sum is taken over all [qr, q2,. . . , q,] in (1,2,. . . ,r~)~-‘. Therefore, for all 1 Q k Q m, tk Q C~=lakx$~)xqz. But ziq = 0 for all 1~ 4 < n because i > 2. Therefore akriq(k) = 0 , by (81, and hence tk = 0 for all 1~ k Q m. Consequently

xi1 = 0.

172

LEROY B. BEASLEY AND NORMAN J. PULLMAN We can show

that

xzj = Q for all j > 2 by the following

similar

argument.

(9) to express xzj as a sum of m terms sk = akx$‘xzj. Then sk < Ctzlakxlqx$) for all k > 1. The rest follows from the fact that zqj = 0 and hence akx$)=Oforalll2. Since zzl = 1, a symmetric argument shows that xlj = 0 for all j > 2 and xi2 = 0 for all i > 2. Therefore, X = B @ C for some 2 X 2 matrix B, completing the proof. Use

REFERENCES J. V. Brawley, Polynomials over a ring that permute the matrices over that ring, J. Algebra 38:93-99 (1976). J. V. Brawley, The number of polynomial functions which permute the matrices over a finite field, J. Cot&in. Theory 21:147-154 (1976). J. V. Brawley, L. Carlitz, and T. Vaughan, Linear permutation polynomials with coefficients in a subfield, Acta A&h. 24:193-199 (1973). J. V. Brawley and R. Gilmer, Fields that admit a nonlinear permutation polynomial, /. Algebra 123:111-119 (1989). J. V. Brawley and G. E. Schnibben, Polynomials which permute the matrices over a field, Linear Algebra Appl. 86:145-160 (1987). C.-L. Chuang, On ranges of polynomials in finite matrix rings, Proc. Amer. Math. sot. 110:293-302 (1990). D. A. Gregory and N. J. Pullman, Semiring rank: Boolelan rank and nonegative factorizations; 1. Combin. Inform. System. Sci. 8:223-233 (1983). K. H. Kim, Boolean Matrix Theory and Appkxztions, Pure Appl. Math. 70, Marcel Dekker, New York, 1982. Receioed 22 October 19W;final

manuscript accepted 17]anuary

1991